随笔分类 -  web前端

ajax请求
摘要：post请求 var url = this.server + "?action=dwm_canbeoperated&tablename=" + tableName; url = $yw.g.id.appendUrl(url); data = JSON.stringify(data); var postData = {"data": data}; $.ajax({ async: f... 阅读全文
posted @ 2017-07-24 11:53 学无止境、 阅读(94) 评论(0) 推荐(0)